Embibe Experts Solutions for Exercise 6: Assignment Set-2

If $\overrightarrow{r}=x(\overrightarrow{a}\times \overrightarrow{b})+y(\overrightarrow{b}\times \overrightarrow{c})+z(\overrightarrow{c}\times \overrightarrow{a})$ and $\left[\begin{array}{lll}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]=\frac{1}{3},$ then $x+y+z$ is equal to

If the vertices of a tetrahedron have the position vectors $\overrightarrow{0},\hat{i}+\hat{j},2\hat{j}-\hat{k}$ and $\hat{i}+\hat{k}$ then the volume of the tetrahedron is

If $\left[\right(2\overrightarrow{a}+\overrightarrow{b}\left)\overrightarrow{c}\overrightarrow{d}\right]=\lambda \left[\overrightarrow{a}\overrightarrow{c}\overrightarrow{d}\right]+\mu \left[\overrightarrow{b}\overrightarrow{c}\overrightarrow{d}\right]$ then $\lambda +\mu$ is equal to

If $\overrightarrow{p}=2\overrightarrow{a}-3\overrightarrow{b},\overrightarrow{q}=\overrightarrow{a}-2\overrightarrow{b}+\overrightarrow{c},\overrightarrow{r}=-3\overrightarrow{a}+\overrightarrow{b}+2\overrightarrow{c},\overrightarrow{a},$ $\overrightarrow{b},\overrightarrow{c}$ being non-zero, non-coplanar vectors, then $-2\overrightarrow{a}+3\overrightarrow{b}-\overrightarrow{c}$ is equal to

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three non-coplanar vectors such that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\alpha \overrightarrow{d}$ and $\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}=\beta \overrightarrow{a}$ then $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}$ is equal to

The position vector of three points are $2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}, \overrightarrow{a}-2\overrightarrow{b}+\lambda \overrightarrow{c}$ and $\mu \overrightarrow{a}-5\overrightarrow{b}$ where $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are non coplanar vectors. The points are collinear when

Let $\overrightarrow{a}=2\hat{i}+2\hat{j}+\hat{k}$ and $\overrightarrow{b}$ be another vector such that $\overrightarrow{a}·\overrightarrow{b}=14$ and $\overrightarrow{a}\times \overrightarrow{b}=3\hat{i}+\hat{j}-8\hat{k},$ then $\overrightarrow{b}=b_1\hat{i}+\hat{j}+b_2\hat{k}$ then $\left(b_1^{b_2}\right)$ is equal to

$P$ be any point on the circumcircle of the equilateral triangle of side length $6$ units, then the value of $|\overrightarrow{PA}{|}^2+|\overrightarrow{PB}{|}^2+|\overrightarrow{PC}{|}^2$ is equal to